How is the integral of $int x / ( sinh(x) + cosh(x) ) dx$ ?

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Answer 1 · 2018-05-16T18:49:38

$-xe^(-x)-e^(-x)+C$

$int x/(sinhx+coshx)*dx$

= $int x/(e^x)*dx$

= $int xe^(-x)*dx$

= $x*(-e^(-x))-int (-e^(-x))*dx$

= $-xe^(-x)+int e^(-x)*dx$

= $-xe^(-x)-e^(-x)+C$

Answer 2 · 2018-05-16T18:52:41

Given: $int x / ( sinh(x) + cosh(x) ) dx$

Multiply the integrand by 1 in the form of $(sinh(x)-cosh(x))/(sinh(x)-cosh(x))$ :

$int x / ( sinh(x) + cosh(x) ) dx = int x / ( sinh(x) + cosh(x) )(sinh(x)-cosh(x))/(sinh(x)-cosh(x)) dx$

The denominator is multiplied using the difference of two squares pattern and the numerator is multiplied using the distributive property:

$int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / ( sinh^2(x) - cosh^2(x)) dx$

The identity $cosh^2(x)- sinh^2(x)=1$ tells us that the denominator is -1:

$int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / -1 dx$

Eliminate the denominator by changing signs in the numerator:

$int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x)-xsinh(x) dx$

Separate into two integrals:

$int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x) dx- int xsinh(x) dx$

Both integrals are trivial integrations by parts:

$int x / ( sinh(x) + cosh(x) ) dx = (xsinh(x) - cosh(x)) - (xcosh(x)-sinh(x))+C$

Regroup:

$int x / ( sinh(x) + cosh(x) ) dx = xsinh(x)- xcosh(x)+ sinh(x)- cosh(x) +C$

How is the integral of int x / ( sinh(x) + cosh(x) ) dxint x / ( sinh(x) + cosh(x) ) dx?